Counting
Techniques for counting
Rule 1
Suppose we have a collection of sets A1, A2, A3,
… and that any pair are mutually exclusive
(i.e. A1 A2 = f) Let
ni = n (Ai) = the number of elements in Ai.
Then N = n( A ) = the number of elements in A
= n1 + n2 + n3 + …
Let A = A1 A2 A3 ….
n1
A1
n2
A2
n3
A3
n4
A4
Rule 2
Suppose we carry out two operations in sequence
Let
n1 = the number of ways the first
operation can be performed
n2 = the number of ways the second
operation can be performed once the
first operation has been completed.
Then N = n1 n2 = the number of ways the two
operations can be performed in sequence.
1n
2n
2n
2n
2n
2n
Diagram:
Examples 1. We have a committee of 10 people. We
choose from this committee, a chairman and
a vice chairman. How may ways can this be
done?
Solution:
Let n1 = the number of ways the chairman can be
chosen = 10.
Let n2 = the number of ways the vice-chairman
can be chosen once the chair has been
chosen = 9.
Then N = n1n2 = (10)(9) = 90
2. In Black Jack you are dealt 2 cards. What is
the probability that you will be dealt a 21?
Solution:
The number of ways that two cards can be selected from
a deck of 52 is N = (52)(51) = 2652.
A “21” can occur if the first card is an ace and the
second card is a face card or a ten {10, J, Q, K} or the
first card is a face card or a ten and the second card is an
ace.
The number of such hands is (4)(16) +(16)(4) =128
Thus the probability of a “21” = 128/2652 = 32/663
Rule 3
Suppose we carry out k operations in sequence
Let
n1 = the number of ways the first operation
can be performed
ni = the number of ways the ith operation can be
performed once the first (i - 1) operations
have been completed. i = 2, 3, … , k
Then N = n1n2 … nk = the number of ways the
k operations can be performed in sequence.
1n
2nDiagram: 3n
2n
2n
Examples 1. Permutations: How many ways can you order n
objects Solution:
Ordering n objects is equivalent to performing n operations in
sequence.
1. Choosing the first object in the sequence (n1 = n)
2. Choosing the 2nd object in the sequence (n2 = n -1).
…
k. Choosing the kth object in the sequence (nk = n – k + 1)
…
n. Choosing the nth object in the sequence (nn = 1)
The total number of ways this can be done is:
N = n(n – 1)…(n – k + 1)…(3)(2)(1) = n!
Example How many ways can you order the 4 objects
{A, B, C, D}
Solution:
N = 4! = 4(3)(2)(1) = 24
Here are the orderings.
ABCD ABDC ACBD ACDB ADBC ADCB
BACD BADC BCAD BCDA BDAC BDCA
CABD CADB CBAD CBDA CDAB CDBA
DABC DACB DBAC DBCA DCAB DCBA
Examples - continued 2. Permutations of size k (< n): How many ways can you
choose k objects from n objects in a specific order
Solution:This operation is equivalent to performing k operations
in sequence.
1. Choosing the first object in the sequence (n1 = n)
2. Choosing the 2nd object in the sequence (n2 = n -1).
…
k. Choosing the kth object in the sequence (nk = n – k + 1)
The total number of ways this can be done is:
N = n(n – 1)…(n – k + 1) = n!/ (n – k)!
This number is denoted by the symbol
!= 1 1
!n k
nP n n n k
n k
Definition: 0! = 1
!= 1 1
!n k
nP n n n k
n k
This definition is consistent with
for k = n
!1
!
!0
!n
nnPnn
Example How many permutations of size 3 can be found in
the group of 5 objects {A, B, C, D, E}
Solution:
5 3
5!= 5 4 3 60
5 3 !P
ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE
ACB ADB AEB ADC AEC AED BDC BEC BED CED
BAC BAD BAE CAD CAE DAE CBD CBE DBE DCE
BCA BDA BEA CDA CEA DEA CDB CEB DEB DEC
CAB DAB EAB DAC EAC EAD DBC EBC EBD ECD
CAB DBA EBA DCA ECA EDA DCB ECB EDB EDC
Example We have a committee of n = 10 people and we
want to choose a chairperson, a vice-chairperson and a
treasurer
Solution: Essentually we want to select 3 persons from the
committee of 10 in a specific order. (Permutations of size 3
from a group of 10).
10 3
10! 10!= 10 9 8 720
10 3 ! 7!P
Solution: Again we want to select 3 persons from the
committee of 10 in a specific order. (Permutations of size 3
from a group of 10).The total number of ways that this can be
done is:
10 3
10! 10!= 10 9 8 720
10 3 ! 7!P
Example We have a committee of n = 10 people and we want
to choose a chairperson, a vice-chairperson and a treasurer.
Suppose that 6 of the members of the committee are male and 4
of the members are female. What is the probability that the
three executives selected are all male?
This is the size, N = n(S), of the sample space S. Assume all
outcomes in the sample space are equally likely.
Let E be the event that all three executives are male
6 3
6! 6!= 6 5 4 120
6 3 ! 3!n E P
Hence
Thus if all candidates are equally likely to be selected to any
position on the executive then the probability of selecting an all
male executive is:
120 1
720 6
n EP E
n S
1
6
Examples - continued 3. Combinations of size k ( ≤ n): A combination of size k
chosen from n objects is a subset of size k where the order of selection is irrelevant. How many ways can you choose a combination of size k objects from n objects (order of selection is irrelevant)
{A,B,C} {A,B,D} { A,B,E} {A,C,D} {A,C,E}
{A,D,E} {B,C,D} {B,C,E} {B,D,E} {C,D,E}
Here are the combinations of size 3 selected from the 5 objects {A, B, C, D, E}
Important Notes
1. In combinations ordering is irrelevant.
Different orderings result in the same
combination.
2. In permutations order is relevant. Different
orderings result in the different permutations.
How many ways can you choose a combination of size k
objects from n objects (order of selection is irrelevant)
Solution: Let n1 denote the number of combinations of size k.
One can construct a permutation of size k by:
1
!Thus !
!n k
nP n k
n k
1. Choosing a combination of size k (n1 = unknown)
2. Ordering the elements of the combination to form
a permutation (n2 = k!)
1
!and the # of combinations of size .
! ! !
n kP nn k
k n k k
The number:
or n k
nC
k
1
1 2 1!
! ! ! 1 2 1
n kn n n n kP n
nk n k k k k k
is denoted by the symbol
read “n choose k”
It is the number of ways of choosing k objects from n
objects (order of selection irrelevant).
nCk is also called a binomial coefficient.
It arises when we expand (x + y)n (the binomial
theorem)
The Binomial theorem:
0 1 1 2 2
0 1 2+ +n n n n
n n nx y C x y C x y C x y
0 + + + k n k n
n k n nC x y C x y
0 1 1 2 2+ + + 0 1 2
n n nn n n
x y x y x y
0 + + + k n k nn n
x y x yk n
Proof: The term xkyn - k will arise when we select x from k
of the factors of (x + y)n and select y from the remaining n
– k factors. The no. of ways that this can be done is:
0 1 1 2 2+ + + 0 1 2
n n n nn n n
x y x y x y x y
n
k
Hence there will be terms equal to xkyn - k and n
k
0 + + + k n k nn n
x y x yk n
Pascal’s triangle – a procedure for calculating binomial
coefficients
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
• The two edges of Pascal’s triangle contain 1’s
• The interior entries are the sum of the two
nearest entries in the row above
• The entries in the nth row of Pascals triangle
are the values of the binomial coefficients
0
n 1
n 3
n 4
n
n
k
n
n
1
n
n
Pascal’s triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1
k
2
k
3
k
4
k
5
k
6
k
7
k
1
x y x y
2 2 2x y x xy y
3 3 2 2 33 3x y x x y xy y
4 4 3 2 2 3 34 6 4x y x x y x y xy y
5 5 4 3 2 2 3 4 55 10 10 5x y x x y x y x y xy y
6 6 5 4 2 3 3 2 4 5 66 15 20 15 6x y x x y x y x y x y xy y
7 7 6 5 2 4 3 3 4 2 5 6 77 21 35 35 21 7x y x x y x y x y x y x y xy y
The Binomial Theorem
Summary of counting results
Rule 1
n(A1 A2 A3 …. ) = n(A1) + n(A2) + n(A3) + …
if the sets A1, A2, A3, … are pairwise mutually exclusive
(i.e. Ai Aj = f)
Rule 2
n1 = the number of ways the first operation can be
performed
n2 = the number of ways the second operation can be
performed once the first operation has been
completed.
N = n1 n2 = the number of ways that two operations can be
performed in sequence if
Rule 3
n1 = the number of ways the first operation can be
performed
ni = the number of ways the ith operation can be
performed once the first (i - 1) operations have
been completed. i = 2, 3, … , k
N = n1n2 … nk
= the number of ways the k operations can be
performed in sequence if
Basic counting formulae
!
!n k
nP
n k
1. Orderings
! the number of ways you can order objectsn n
2. Permutations
The number of ways that you can
choose k objects from n in a
specific order
!
! !n k
n nC
k k n k
3. Combinations
The number of ways that you
can choose k objects from n
(order of selection irrelevant)
Applications to some counting
problems
• The trick is to use the basic counting formulae
together with the Rules
• We will illustrate this with examples
• Counting problems are not easy. The more
practice better the techniques
Application to Lotto 6/49
Here you choose 6 numbers from the integers 1,
2, 3, …, 47, 48, 49.
Six winning numbers are chosen together with a
bonus number.
How many choices for the 6 winning numbers
49 6
49 49 48 47 46 45 4449!
6 6!43! 6 5 4 3 2 1C
13,983,816
You can lose and win in several ways
1. No winning numbers – lose
2. One winning number – lose
3. Two winning numbers - lose
4. Two + bonus – win $5.00
5. Three winning numbers – win $10.00
6. Four winning numbers – win approx. $80.00
7. 5 winning numbers – win approx. $2,500.00
8. 5 winning numbers + bonus – win approx. $100,000.00
9. 6 winning numbers – win approx. $4,000,000.00
Counting the possibilities
1. No winning numbers – lose
All six of your numbers have to be chosen from the losing numbers
and the bonus
1. One winning number – lose
2. Two winning numbers - lose
3. Two + bonus – win $5.00
4. Three winning numbers – win $10.00
5. Four winning numbers – win approx. $80.00
6. 5 winning numbers – win approx. $2,500.00
7. 5 winning numbers + bonus – win approx. $100,000.00
8. 6 winning numbers – win approx. $4,000,000.00
Counting the possibilities
2. One winning numbers – lose
All six of your numbers have to be chosen from the losing numbers
and the bonus.
43 6,096,454
6
1. No winning numbers – lose
One number is chosen from the six winning numbers and the
remaining five have to be chosen from the losing numbers and the
bonus.
6 43
6 962,598 = 5,775,588 1 5
4. Two winning numbers + the bonus – win $5.00
Two numbers are chosen from the six winning numbers and the
remaining four have to be chosen from the losing numbers (bonus
not included)
3. Two winning numbers – lose
Two numbers are chosen from the six winning numbers, the
bonus number is chose and the remaining three have to be chosen
from the losing numbers.
6 1 42
15 1 11,480 = 172,2002 1 3
6 42
15 111,930 = 1,678,9502 4
6. four winning numbers – win approx. $80.00
Three numbers are chosen from the six winning numbers and the
remaining three have to be chosen from the losing numbers + the
bonus number
5. Three winning numbers – win $10.00
Four numbers are chosen from the six winning numbers and the
remaining two have to be chosen from the losing numbers + the
bonus number
6 43
15 903 = 13,5454 2
6 43
20 12,341 = 246,8203 3
8. five winning numbers + bonus – win approx. $100,000.00
Five numbers are chosen from the six winning numbers and the
remaining number has to be chosen from the losing numbers
(excluding the bonus number)
7. five winning numbers (no bonus) – win approx. $2,500.00
Five numbers are chosen from the six winning numbers and the
remaining number is chosen to be the bonus number
6 1
6 1 = 65 1
6 42
6 42 = 2525 1
Six numbers are chosen from the six winning numbers,
9. six winning numbers (no bonus) – win approx. $4,000,000.00
6 1
6
Summary
n Prize Prob
0 winning 6,096,454 nil 0.4359649755
1 winning 5,775,588 nil 0.4130194505
2 winning 1,678,950 nil 0.1200637937
2 + bonus 172,200 5.00$ 0.0123142353
3 winning 246,820 10.00$ 0.0176504039
4 winning 13,545 80.00$ 0.0009686197
5 winning 252 2,500.00$ 0.0000180208
5 + bonus 6 100,000.00$ 0.0000004291
6 winning 1 4,000,000.00$ 0.0000000715
Total 13,983,816
Another Example counting poker hands
A poker hand consists of five cards chosen at
random from a deck of 52 cards.
The total number of poker hands is
522,598,960
5N
6
6
AAA
AA
A
A
A6
6
6
6
6
6
AA
AA
AAA
AA
AA
AA
AA
AA
AA
A
A
A
A
A
A
A
Types of poker hand counting poker hands
1. Nothing Hand {x, y, z, u, v}
• Not all in sequence or not all the same suit
2. Pair {x, x, y, z, u}
3. Two pair {x, x, y, y, z}
4. Three of a kind {x, x, x, y, z}
5. Straight {x, x+ 1, x + 2, x + 3, x + 4}
• 5 cards in sequence
• Not all the same suit
6. Flush {x, y, z, u, v}
• Not all in sequence but all the same suit
7. Full House {x, x, x, y, y}
8. Four of a kind {x, x, x, x, y}
9. Straight Flush {x, x+ 1, x + 2, x + 3, x + 4}
• 5 cards in sequence but not {10, J, Q, K, A}
• all the same suit
10. Royal Flush {10, J, Q, K, A}
• all the same suit
counting the hands 2. Pair {x, x, y, z, u}
3. Two pair {x, x, y, y, z}
We have to:
• Choose the value of x
• Select the suits for the for x.
• Choose the denominations {y, z, u}
• Choose the suits for {y, z, u} - 4×4×4 = 64
13 13
1
4
62
12
2203
Total # of hands of this type = 13 × 6 × 220 × 64 = 1,098,240
• Choose the values of x, y
• Select the suits for the for x and y.
• Choose the denomination z
• Choose the suit for z - 4
13 78
2
4 4
362 2
11
111
Total # of hands of this type = 78 × 36 × 11 × 4 = 123,552
We have to:
4. Three of a kind {x, x, x, y, z}
• Choose the value of x
• Select the suits for the for x.
• Choose the denominations {y, z}
• Choose the suits for {y, z} - 4×4 = 16
13 13
1
4
43
12
662
Total # of hands of this type = 13 × 4 × 66 × 16 = 54,912
We have to:
7. Full House {x, x, x, y, y}
• Choose the value of x then y
• Select the suits for the for x.
• Select the suits for the for y.
13 2 13 12 156P
4 4
3
4
62
We have to:
Total # of hands of this type = 156 × 4 × 6 = 3,696
• Choose the value of x
• Select the suits for the for x.
• Choose the denomination of y.
• Choose the suit for y - 4
13 13
1
4
14
12
121
Total # of hands of this type = 13 × 1 × 12 × 4 = 624
We have to:
8. Four of a kind {x, x, x, x, y}
9. Straight Flush {x, x+ 1, x + 2, x + 3, x + 4}
• 5 cards in sequence but not {10, J, Q, K, A}
• all the same suit
10. Royal Flush {10, J, Q, K, A}
• all the same suit
Total # of hands of this type = 9×4 = 36 (no. of suits)
Total # of hands of this type = 4 (no. of suits)
The hand could start with {A, 2, 3, 4, 5, 6, 7, 8, 9}
5. Straight {x, x+ 1, x + 2, x + 3, x + 4}
• 5 cards in sequence
• Not all the same suit
• Choose the starting value of the sequence, x.
Total of 10 possibilities {A, 2, 3, 4, 5, 6, 7,
8, 9, 10}
• Choose the suit for each card
4 × 4 × 4 × 4 × 4 = 1024
We have to:
Total # of hands of this type = 1024 × 10 - 36 - 4 = 10200
We would have also counted straight flushes
and royal flushes that have to be removed
6. Flush {x, y, z, u, v}
• Not all in sequence but all the same suit
• Choose the suit 4 choices
• Choose the denominations {x, y, z, u, v}
We have to:
Total # of hands of this type = 1287 × 4 - 36 - 4 = 5108
We would have also counted straight flushes
and royal flushes that have to be removed
131287
5
Summary
Frequency Prob.
nothing 1,302,588 0.50119586
pair 1,098,240 0.42256903
two pair 123,552 0.04753902
3 of a kind 54,912 0.02112845
straight 10,200 0.00392465
flush 5,108 0.00196540
full house 3,696 0.00142211
4 of a kind 624 0.00024010
straight flush 36 0.00001385
royal flush 4 0.00000154
Total 2,598,960 1.00000000
Next Topic: Probability Rules